If A= [ array ll 0 & 1 0 & 0 array ] , then (a I+b A)^n is (where I is the identify matrix of order 2)

Step-by-step solution

Given, A= [ array ll 0 & 1 0 & 0 array ] Let P(n):(a I+b A)^n=a^n I+n a^ n-1 b A For n=1, p(1):(a I+b A)^1=a^1 I+1 a¹⁻¹ b A a I+b A=a I+b A It is true for n=1 . Let P(n) is true for n=K . So, P(K):(a I+b A)^K=a^K I+K a^ K-1 b A Now, for n=K+1 aligned & P(K+1): & (a I+b A)^ K+1 =a^ K+1 I+(K+1) a^ K+1-1 b A & LHS =(a I+b A)^ K+1 =(a I+b A)^K(a I+b A) & = (a^K I+K a^ K-1 b A )(a I+b A) [from Eq. (i)] & =a^ K+1 I+K a^K I b A+a^K I b A+K a^ K-1 b^2 A^2 & =a^ K+1 I+K a^K b(I A)+a^K b(I A)+K a^ K-1 b^2 (A^2 ) & =a^ K+1 I+